Question

The simply-supported beam having I-beam cross-section as shown in figure is to carry a uniformly distributed load over its entire 1.2m length. Specify the maximum allowable load if the beam is made from malleable iron, ASTM A220, class 80002. The allowable tensile stress is 164 MPa and allowable compressive stress is 412 MPa. The centroid of the section is located at 35 mm from the bottom and moment of inertia are Ix = 2.66 x 10 mm". (a) Draw loading diagram and compute support reactions in terms of w (intensity of UDL). (6) (b) Draw shear force diagram. (c) Compute maximum bending moment in terms of w) (d) Draw bending moment diagram. The BMD will explain whether beam is sagging or, hogging. (4) (e) Compute w for top fiber and w for bottom fiber. Then specify the safe w and provide (5) reasons. 40 30 Loading Diagram 35 mm 3 am

Answer 1

Given

$\\ I_x=2.66*10^5 \ mm^4 \\ \sigma_{allow,tensile}=164\ MPa \\ \sigma_{allow,comp}=412\ MPa \\ L=1.2\ m \\$

a)Loading diagram

w N/m X A B L RB RA

Taking moment Balance about A

Taking force balance

RBL = WL* L/2 RB = wL/2

RA+ RB = WL RA = WL/2

b)

Shear force diagram

WL/2 B A 0 L L/2 -WL/2

c)

$\\ At\ \frac{\mathrm{d} M}{\mathrm{d} x}=0\ \\ \frac{\mathrm{d} ^2M}{\mathrm{d} x^2}\ =-w\ (it\ is \ nagative) \\ So\ at\ \frac{\mathrm{d} M}{\mathrm{d} x}=0\ bending \ moment \ is\ maximum \\ \frac{\mathrm{d} M}{\mathrm{d} x}=V(shear\ force)$

At mid-span shear force is zero so the bending moment is maximum at mid-span

d)

The bending moment at A and B is zero and the maximum bending moment at mid-span.

Since the second derivative of the bending moment is constant the bending diagram should be a parabola.

Bending moment diagram

wL318 В A L/2

Since bending moment diagram have positive bending moment throughout

Therefore the beam will be sagging

e)

Maximum bending moment

Bending stress equation

$\\ \sigma=-\frac{My}{I_x}$

At mid-span, we will get maximum bending stress

40 OS 35 mm 20

$\\ For\ top \ fiber \\ \sigma_{top\ fiber}=-\frac{M_{max}*(50+5+5-35)}{I_x}=-\frac{125wL^2*25}{I_x}=- 0.016917z(compressive ) \\ Since\ it \ is \ compressive \\ z_{allow,top}=\frac{\sigma_{allow,comp}}{ 0.016917}= 24353.778\ N/m=24.354\ kN/m$

$\\ For\ bottom \ fiber \\ \sigma_{bottom\ fiber}=-\frac{M_{max}*-35}{I_x}=\frac{125wL^2*35}{I_x}= 0.023684z(tensile ) \\ Since\ it \ is \ compressive \\ z_{allow,bottom}=\frac{\sigma_{allow,tensile}}{ 0.023684}=6924.4\ N/m=6.924\ kN/m$

The allowable loading for bottom fiber is small so it is the safest loading

If we chose the allowable loading for top fiber then the bottom fiber will fail.

So the safest maximum loading is the 6.924 kN/m